Answer:
Initial concentration of HI is 5 mol/L.
The concentration of HI after 
is 0.00345 mol/L.
Explanation:
Rate Law: ![k[HI]^2 ](https://tex.z-dn.net/?f=k%5BHI%5D%5E2%0A)
Rate constant of the reaction = k = 
Order of the reaction = 2
Initial rate of reaction = 
Initial concentration of HI =![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
![1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2](https://tex.z-dn.net/?f=1.6%5Ctimes%2010%5E%7B-7%7D%20mol%2FL%20s%3D%286.4%5Ctimes%2010%5E%7B-9%7D%20L%2Fmol%20s%29%5BHI%5D%5E2)
![[A_o]=5 mol/L](https://tex.z-dn.net/?f=%5BA_o%5D%3D5%20mol%2FL)
Final concentration of HI after t = [A]
t =
Integrated rate law for second order kinetics is given by:
![\frac{1}{[A]}=kt+\frac{1}{[A_o]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3Dkt%2B%5Cfrac%7B1%7D%7B%5BA_o%5D%7D)
![\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3D6.4%5Ctimes%2010%5E%7B-9%7D%20L%2Fmol%20s%5Ctimes%204.53%5Ctimes%2010%5E%7B10%7D%20s%2B%5Cfrac%7B1%7D%7B%5B5%20mol%2FL%5D%7D)
![[A]=0.00345 mol/L](https://tex.z-dn.net/?f=%5BA%5D%3D0.00345%20mol%2FL)
The concentration of HI after is 0.00345 mol/L.
Answer:
The answer is Sodium Sulfate = Na2SO4
Explanation:
Molar mass of sulfate = 1 (S) + 4 (O) = 1 (32) + 4 (16) = 32 + 64 = 96
Molar mass of sodium sulfate = 2 (23) + 96 = 46 + 96 = 142
% of Sulfate = (96/142)*100 = 67.6%
Percent mistake in Studen A,
(I) % mistake = (67.6 - 68.6)/67.6 = 1.48
(ii) % mistake = (67.6 - 66.2)/67.6 = 2.07
(iii) % mistake = (67.6 - 67.1)/67.6 = 0.74
For understudy B
(I) % mistake = (67.6 - 66.7)/67.6 = 1.33
(ii) % mistake = (67.6 - 66.6)/67.6 = 1.48
(iii) % mistake = (67.6 - 66.5)/67.6 = 1.63
Sutdent An is some how exact.
Understudy B is exact however not precise.
The solution for problem 36 is in the picture attached. I am 80% sure it’s correct.
-NH2 is the most favorable for the reaction
